Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
#include<iostream>
#include<sstream>
#include<algorithm>
#include<climits>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<ctime>
#include<cfloat>
#include<functional>
#include<map>
#include<string>
#include<cstring>
#include<vector>
#include<queue>
#include<stack>
#include<deque>
#include<set>
#include<bitset>
#include<list>
#include<numeric>
#include<complex>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> i_i;
typedef pair<long long, int> ll_i;
typedef pair<double, int> d_i;
typedef pair<long long, long long> ll_ll;
typedef pair<double, double> d_d;
typedef vector<int> Vint;

#define PI 3.141592653589793238462643383279
#define mod 1000000007LL
#define rep(i, n) for(i = 0;i < n;++i)
#define rep1(i, n) for(i = 1;i < n;++i)
#define rep2d(i, j, n) for(i = 0;i < n;++i)for(j = i + 1;j < n;++j)
#define per(i, n) for(i = n - 1;i > -1;--i)
#define int(x) int x; scanf("%d",&x)
#define int2(x, y) int x, y; scanf("%d%d",&x, &y)
#define int3(x, y, z) int x, y, z; scanf("%d%d%d",&x, &y, &z)
#define int4(v, x, y, z) int v, x, y, z; scanf("%d%d%d%d", &v, &x, &y, &z)
#define int5(v, w, x, y, z) int v, w, x, y, z; scanf("%d%d%d%d%d", &v, &w, &x, &y, &z)
#define scn(n, a) rep(i, n)cin >> a[i]
#define sc2n(n, a, b) rep(i, n)cin >> a[i] >> b[i]
#define pri(x) cout << x << "\n"
#define pri2(x, y) cout << x << " " << y << "\n"
#define pri3(x, y, z) cout << x << " " << y << " " << z << "\n"
#define pb push_back
#define mp make_pair
#define all(a) (a).begin(),(a).end()
#define endl "\n"
#define kabe puts("---------------------------")
#define kara puts("")
#define debug(x) cout << " --- " << x << "\n"
#define debug2(x, y) cout << " --- " << x << " " << y << "\n"
#define debug3(x, y, z) cout << " --- " << x << " " << y << " " << z << "\n"
#define X first
#define Y second
#define eps 0.0001
#define prid(x) printf("%.15lf\n", x)

struct edge{
  int to, cost;
};
int qwe = 0;
vector<edge> G[100000];
ll d[2][100000];
ll lis[2][100000] = {0};

ll longinf = 123451234512345LL;

void dijkstra(int v, int e, int r){   //(vertex, edge, start)
  ll res = 0;
  int i, j;
  rep(i, v)d[qwe][i] = longinf;
  d[qwe][r] = 0LL;
  priority_queue<ll_i, vector<ll_i>, greater<ll_i> > Q;
  Q.push(mp(0LL, r));
  for(;!Q.empty();){
    ll_i now = Q.top();
    Q.pop();
    if(d[qwe][now.Y] < now.X)continue;
    rep(i, G[now.Y].size()){
      edge tmp = G[now.Y][i];
      if(d[qwe][tmp.to] > d[qwe][now.Y] + tmp.cost){
        d[qwe][tmp.to] = d[qwe][now.Y] + tmp.cost;
        Q.push(mp(d[qwe][tmp.to], tmp.to));
      }
    }
  }
  return;
}

signed main(void){
  int i, j, k;
  for(int testcase = 0;testcase < 1234567;testcase++){
    int4(n, m, s, t);
    --s; --t;
    rep(i, m){
      int2(x, y);
      --x; --y;
      G[x].pb((edge){y, 1});
      G[y].pb((edge){x, 1});
    }

    dijkstra(n, m * 2, s); qwe++;
    dijkstra(n, m * 2, t);
    int dist = d[0][t];

    rep(i, 2)rep(j, n)if(d[i][j] < longinf)lis[i][d[i][j]]++;

    ll res = 0;
    rep(i, dist - 1)res += lis[0][i] * lis[1][dist - 2 - i];
    pri(res);


/*/

//*/ break;
  }
  return 0;
}